Final answer:
The voltage across a 10mH inductor with current I=10e^{-t/2} A is V=-50mV * e^{-t/2}, and the power dissipated is P=-500mW * e^{-t}. Both calculations involve determining the time rate of change of current and utilizing the relationship between voltage, current, and inductance.
Step-by-step explanation:
To find the voltage across the inductor, we can use the formula for the induced emf in an inductor, which is given by V=L(di/dt), where V is the voltage, L is the inductance, and di/dt is the time rate of change of current. Since we are given the current through a 10mH inductor as I=10e^{-t/2} A, we can find di/dt by differentiating I with respect to t, which gives us di/dt=-5e^{-t/2} A/s. Substituting the values, we get V=10mH * (-5e^{-t/2} A/s) = -50mV * e^{-t/2}.
To find the power, we use the formula P=VI, where P is the power, V is the voltage, and I is the current. By substituting the expressions for V and I, we get P=(-50mV * e^{-t/2})(10e^{-t/2} A) = -500mW * e^{-t}, which is the power dissipated in the inductor.